Question

In an attempt to escape from a deserted island, a castaway builds a raft and sets sail for the sea. The wind changes a lot during the day and flies along the following straight lines: 2.60 km and 45.0 ° north of west, then 4.80 km and 60.0 ° south of east, then 1.60 km and 5.00 ° east of north, and finally 2.70 km and 13.0 ° north of the east. Use the analytical method to find the resulting vector of all its displacement vectors. What is its magnitude and direction?

Answer 1

Let´s analyze each movement and find its perpendicular components. Then, we are going to add the components along the x and y axes to get the resultant and finally to calculate its magnitude and direction.

To determine the perpendicular components, we will use right triangles trigonometrics ratios.

• Displacement A: 2.60 km and 45° north of west

Ax= AcosФ = 2.60cos45° = 1.84 km (neg. direction)

Ay= AsinФ = 2.60sin45° = 1.84 km (pos. direction)

• Displacement B: 4.80 km and 60° south of east

Bx= BcosФ = 4.80cos60° = 2.40 km (pos. direction)

By= BsinФ = 4.80sin60° = 4.15 km (neg. direction)

• Displacement C: 1.60 km and 5° east of north

Cx= CsinФ = 1.60sin5° = 0,14 km (pos. direction)

Cy= CcosФ = 1.60cos5° = 1.59 km (pos. direction)

• Displacement D: 2.70 km and 13° north of east

Dx= DcosФ = 2.70cos13° = 2.63 km (pos. direction)

Dy= DsinФ = 2.70sin13° = 0.60 km (pos. direction)

Now, we add the components along the x- and y- axis to find the components of the resultant (R):

Rx = Ax + Bx + Cx + Dx

Ry = Ay + By + Cy + Dy

Therefore,

Rx=(-1.84)+(2.4)+(0.14)+(2.63)

Ry=(1.84)+(-4.15)+(1.59)+(0.60)

Rx=3.33

Ry=-0.12

Note: minus (-) symbol to negative directions

Let´s use the Theorem of Pythagoras to find the magnitud ot the resultant:

R=3.332km

To find the direction of the resultant:

tanФ =

tanФ =

Ф=2.06 degree north of east